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Integral (cos x)^2 dxby vadik
Tags: integral 
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#1
Apr2804, 04:36 PM

P: 23

Do any one have an idea how to calculate integral of (cos x)^2 ? Or is it even possible? I tried some substitutions and/or rules of trigonometry, like cosxcosx+sinxsinx=1, but it didn't help. Thank you!



#2
Apr2804, 04:42 PM

Sci Advisor
P: 6,039

cos^{2}x+sin^{2}x=1
cos^{2}xsin^{2}x=cos2x Therefore cos^{2}x=(1+cos2x)/2 I'll let you finish. 


#3
Apr2804, 04:44 PM

P: 23

Thank you. :) integral (cos x)^2 dx



#4
Mar308, 10:12 AM

P: 14

Integral (cos x)^2 dx
dont you have to use half angle identities to get integral of cos^2 ?



#5
Mar308, 12:00 PM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,345

No, double angle formulas as mathman said.



#6
Mar508, 03:01 PM

P: 169

an easy way to remember the solution to this common integral, when integrating over a whole period:
cos^2 x + sin ^2 x =1 [tex] \int cos^2 x = \int sin^2 x [/tex] , at least when you integrate over a whole period [tex] \int cos^2 x + \int sin^2 x =[/tex] length of a period so the integral gives length of a period divided by 2 


#7
Mar508, 06:59 PM

P: 188

Why does this thread have over 16,000 views?
edit: Oh, it's four years old. 


#8
Sep1309, 12:20 AM

P: 1

First use the halfangle formula to change the cos(x)^2 to (1+cos(2x))/2...
This will allow you to break the integral into two seperate problems much easier to solve integral{ 1/2dx + integral{ cos(2x)dx Then you will have x/2 + (sin(2x)/2) + C 


#9
Sep1309, 01:24 AM

P: 1,105

What the, that's not even correct. If you're gonna revive a 5year old thread, at least make sure you don't have arithmetic errors.



#10
Sep1409, 11:55 PM

P: 79

sin(2x)/4 ;)



#11
Feb2210, 05:35 AM

P: 1

use the euler's formula
cos x= [e^ix+e^ix ] [] [ 2 ] 


#12
Feb2210, 06:20 AM

P: 532

http://www.5min.com/Video/AnIntrodu...sine169056088
Why doesn't the student, after nearly 6 years of unsuccessfully attempting this crazy integral, try a visual aid? 


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